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(B) Given $\vec{a} \perp \vec{b}$ and $\vec{a} \perp \vec{c}$, we have $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$. Given $|\vec{a}+\v

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$|\vec{a}| |\vec{b}| |\vec{c}|$

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(B) Given $\vec{a} \perp \vec{b}$ and $\vec{a} \perp \vec{c}$, we have $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$. Given $|\vec{a}+\vec{b}+\vec{c}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}$. Squaring both sides: $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2$. Since $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$, this simplifies to $2(\vec{b} \cdot \vec{c}) = 0$, so $\vec{b} \cdot \vec{c} = 0$. Now, consider the vector triple product: $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$. Since $\vec{a} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{c} = 0$, we get $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = \vec{0}$, so $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = 0$. Next, the scalar triple product is $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |[\vec{a} \vec{b} \vec{c}]|$. Since $\vec{a} \perp \vec{b}$ and $\vec{a} \perp \vec{c}$ and $\vec{b} \perp \vec{c}$, the vectors $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular. Thus, $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}|$. Therefore, $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}|+|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}| + 0 = |\vec{a}| |\vec{b}| |\vec{c}|$.

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